Suppose we have two integers $b$ and $c$.

If $\gcd(b, c) = 1$, then the pair of integers is relatively prime.

$\gcd(b, c) = 1$ can also be expressed as a linear combination $bx_0 + cy_0 =1$, where $x_0$ and $y_0$ are integers chosen such that $bx_0 + cy_0$ is the smallest positive element in the set $ \{bx + cy\} $, where $x$ and $y$ are integers.

From the above, we can conclude that all pairs of consecutive integers are relatively prime pairs considering:

$(1)(x+1) + (-1)(x) = 1 \implies \gcd(x, x+1) = 1$ for an integer $x$

Now, consider the relatively prime pair of numbers $37$ and $39$. Because $\gcd(37, 39) = 1$ we know that there must be some $x_0$ and some $y_0$ for which $37x_0 + 39y_0 = 1$. That means there are two consecutive integers represented by $37x_0$ and $39y_0$. (We also know that a pair of consecutive integers must correspond to every pair of relatively prime numbers that is not consecutive because of equality of the linear combination to 1).

I have found $x_0 = 19, y_0 = 18$ representing the pair of consecutive numbers $\gcd(703, 702)$ that is related to $\gcd(37, 39)$.

I want to better understand the relationship between pairs of relatively prime numbers that are not consecutive numbers and their relatively prime numbers that are consecutive numbers. For example, is there a unique pair of consecutive numbers for every relatively prime set of nonconsecutive numbers?


For your latter question, no.

$$(6)+(-5) = (2){\color{red}{(3)}}+(-1){\color{red}{(5)}}=1$$ $$(-9)+(10) = (-3){\color{red}{(3)}}+(2){\color{red}{(5)}}=1$$

For $(a,b)=1$, then there exist integers $m,n$ with $ma+nb=1$.

By adding and subtracting $ab$, we get a new solution

$$ma+nb +ab - ab = 1 \iff (m+b)a + (n-a)b = 1$$

You can continue this indefinitely.

And so one question you may ask is:

How often can the sequence $\{(|(m+kb)a|, |(n-ka)b|)\}_{k\in\mathbb Z}$ be a consecutive pair?

  • $\begingroup$ Would there be a way to know exactly how many pairs of consecutive numbers are associated with any pair of nonconsecutive numbers? $\endgroup$ – Richard K Yu Dec 1 '19 at 1:39
  • $\begingroup$ Hopefully my edit helps you some $\endgroup$ – David P Dec 1 '19 at 2:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.